How to prove this logarithmic inequality?

[anemone]

Hi all, I've been having a hard time trying to solve the following inequality:

Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$

I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.

Could I get some hints on how to tackle this problem?

Any help would be deeply appreciated.

Thanks!

P.S. This question was originally asked here (http://www.mymathforum.com/viewtopic.php?f=13&t=27644&p=110515&hilit=noki#p110515) at MMF.

 

[CaptainBlack]

Hi all, I've been having a hard time trying to solve the following inequality:

Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$

I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.

Could I get some hints on how to tackle this problem?

Any help would be deeply appreciated.

Thanks!

P.S. This question was originally asked here (http://www.mymathforum.com/viewtopic.php?f=13&t=27644&p=110515&hilit=noki#p110515) at MMF.

\(\log_{24}(48)=1+\log_{24}(2)\)

But \(2^5 \gt 24\) so \(\log_{24}(2) \gt 1/5\)

Also: \(\log_{12}(54)=1+\log_{12}(4.5)\), and \(4.5^5>12^3\) so \(\log_{12}(4.5)>3/5\)

Hence:
\[ (\log_{24}(48))^2 + (\log_{12}(54))^2 \gt 1.2^2+1.6^2 =4 \]

CB

 

[anemone]

Hi CB, a big thank for your help in making it so straightforward and simple for me!

Thanks.